Given that x and y are positive integers, is x prime?

Let's analyze the question.

The question stem just tells us that x and y are positive integers. The information was provided mainly to rule out a decimal value for x and since we are using factorials for y.
Question: Is x prime?

I. \((y + 1)! <= x <= (y + 1)(y! + 1)\)
This needs to be modified a little. Why? because right now, there is no apparent relation between (y+1)! and (y + 1)(y! + 1). (y+1)! makes sense to me, (y! +1) does not. Can I bring everything in terms of (y+1)?
I see that I have a (y+1) multiplied with y! and with 1. Recognize that (y+1)*y! = (y+1)!
So, (y + 1)(y! + 1) = (y+1)y! + (y+1) = (y+1)! + (y+1)
How will you know that this is how you would like to split it? Use (y+1)! on left side as a clue. The right side should make sense with respect to the left side which is in its simplest form.

\((y + 1)! <= x <= (y+1)! + (y+1)\)

Now think about it. x can take any of the following values (general case):
(y+1)! - Not prime except if y is 1
(y+1)! + 1 - Cannot say whether it is prime or not. If y = 1, this is prime. If y is 2, this is prime. If y is 3 it is not.
(y+1)! + 2 - Has 2 as a factor. Not prime
(y+1)! + 3 - Has 3 as a factor. Not prime
.
.
(y+1)! + (y+1) - Has (y+1) as a factor. Not prime

Hence x may or may not be prime.

II. \((y + 1)! + 1\) has five factors
Not sufficient on its own. No mention of x.

Together: There were two exceptions we found above.
1. If y = 1, then (y+1)! is prime.
From stmnt II, since (1+1)! + 1 = 3 does not have 5 factors (It only has 2), y is not 1. Hence y is not 1 and (y+1)! is not prime
2. We don't know whether (y+1)! + 1 is prime
Since (y+1)! + 1 has 5 factors, it is definitely not prime. Prime numbers have only 2 factors.
Since both exceptions have been dealt with using both statements together, we can say that in every case, x is not prime.
Answer (C).

The question is not time consuming if you quickly see the pattern. It's all about getting exposed to the various concepts. The explanation seems long but only because I have written out everything my mind thinks in a few seconds. This is a relatively tough question but not beyond GMAT. Expect such questions if you are shooting for 49-51 in Quant.

(1) implies
(y+1)! <= x <= (y+1)! + (y+1)
Lets replace y+1 by a

=> a! <= x <= a! + a

For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a
=>x is not prime.

(2) No result derivable.

Answer should be A.

OA please.
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Agreed ! 100% it has to be A.
a will have factors between a! + a and a!. x is NOT prime.

I think this pattern is called "LONE WOLF" trap. it should be A.

https://gmatclub.com/wiki/GMAT_math_tips

I now think that above conclusion can also be reached as any number that has more than 2 factors can be written in the form of a^p*b^q.. and so on where a and b are distinct prime numbers and the number of factors for such a number would be given by (p+1)*(q+1). Therefore, any number that has 5 factors would necessarily have the form Z^4 where z is a prime.
However, the second statement in the question is a red herring, as we would get really confused if we try and grapple with the problem of finding a value for y for which (y + 1)! + 1 has five factors. All we need to be concerned here with is the fact that because (y + 1)! + 1 has five factors (y + 1)! + 1 is not prime.

Superb question Karishma +1

Dear Karishma,

Very beautiful problem, congrats!

Let me contribute with a different wording.

I recognize the famous result: for any integer n greater than 1, we don´t have primes in the interval [ n!+2 , n!+n ] (*) .
(Because j is a factor of n!+j , where j is 2,3, ..., n.)

\(x,y\,\, \ge 1\,\,{\rm{ints}}\)

\(x\,\,\mathop {\rm{ = }}\limits^? \,\,{\rm{prime}}\)

\(\left( 1 \right)\,\,\,n!\,\, \le x\,\, \le \,\,n!\, + n\,\,\,,\,\,{\rm{where}}\,\,\,n = y + 1\,\,\,\,\,\,\,\left[ {\,n!\, + n\,\, = \,\,\left( {y + 1} \right) \cdot y!\, + \left( {y + 1} \right) = \left( {y + 1} \right)\left( {y!\, + 1} \right)\,} \right]\)

\(\left\{ \matrix{\\
\,{\rm{Take}}\,\,y = 1\,\,\left( {n = 2} \right)\,\,{\rm{and}}\,\,x\,{\rm{ = }}\,{\rm{2}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,y = 1\,\,\left( {n = 2} \right)\,\,{\rm{and}}\,\,x\,{\rm{ = }}\,4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{\\
\,n!\,\, \le x\,\, \le \,\,n!\, + n \hfill \cr \\
\,n!\,\, + \,\,1\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\, \Rightarrow \,\,n \ge 3\,\,\left( {y \ge 2} \right)\,\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\{ \matrix{\\
\,x = n!\,\,\,{\rm{not}}\,\,{\rm{prime}} \,\,\, (n \ge 3) \hfill \cr \\
\,x = n!\,\, + \,\,1\,\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\left( {{\rm{statement}}\,\,\left( 2 \right)} \right) \hfill \cr \\
\,n!\, + 2\,\, \le x\,\, \le \,\,n!\, + n\,\,\,,\,\,x\,\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle\)

Kind Regards,
Fabio.
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this question is really tough and time-consuming.
my answer is E and it took me > 5 mins to solve it without knowing if i am right.

statement 2 is insufficient cos we dont know x
statement 1:
(y+1)!=1*2*...*y*(y+1) =1.2.....y.(y+1)=
=1.2.....y.y +1.2.3....y=y.y!+y!
(y+1)*(y!+1)=y.y!+y!+y+1=(y+1)!+(y+1)
=> (y+1)!<=x<=(y+1)!+(y+1)
whatever y is, prime or not prime, (y+1)! is never a prime
if y=1 so 2!=2<x<4 > x=3
y=2 so 3!=6<x<6+2+1=9 so x coule be 7,8,9 so insufficient

statement 1+2
(y+1)!+1<=x<=(y+1)!+1+(y+1)
A<=x<=A+y+1 ( consider A to be (y+1)!+1)
insufficient
E

( still confused)

I think the answer is C.
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Looks like this is pretty time consuming. On the D Day, I'll move on to the next question after spending 30 seconds and understanding the complexity.

The answer looks like E.
Experts ?

The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime

From statement 2) can I infer that (y+1)! + 1 is perfect square. some power of 4. ie. z^4 where z > 0.

So far, I am not able to find a single integer z, that satisfies this criteria. I wonder if there is a problem with second statement.
II. (y + 1)! + 1 has five factors

Sorry cant read those y+1s. Lets substitute y+1 with a.
I a! <= x <= a!+a
II a! + 1 has five factors.

1) Insufficient. We dont know if a is prime. We also dont know if a! is prime.

2) Insufficient. a!+1 has five factors. We can infer a! is not prime. And a is not prime. We don't know x.

Combine 1) + 2) Sufficient.

The pattern is [a! + Integer] has atleast 3 factors namely 1, (a! + Integer) and F such that 2 <= F <= a
Here Integer <= a

Agree?

Brilliant ! Makes all the sense now.

Hi Karishma

How is :

(y+1)! + 3 - Has 3 as a factor. Not prime

if y = 1, then (2! +3) = 5, right ?

Regards,
Subhash
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We are considering all terms from (y+1)! to (y+1)! + (y+1)
If y = 1, then we are only considering terms 2!, 2! + 1 and 2! + 2.

I have given the general case above where y is some greater number say 6.
In that case
6!
6! + 1
6! + 2 - factor 2
6! + 3 - factor 3
6! + 4 - factor 4
6! + 5 - factor 5
6! + 6 - factor 6

Experts help me understand the combination part.

Responding to a pm:



You are right that a number with 5 factors will be a perfect square.

(2) \((y + 1)! + 1\) has five factors

Note that we are NOT given that (y + 1)! has 5 factors. We are not given that a factorial has 5 factors. We are given that (y+1)! + 1 has 5 factors.
We are given that 1 more than a factorial is a perfect square. That is possible.

e.g.
4! + 1 = 25 (a perfect square)

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